Pertanyaan

x → 0 lim x tan x cos x − cos 2 x = ...

$x \to 0 lim \frac{cos x - cos 2 x}{x tan x} = ...$

P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Ingatlah rumus selisih cosinus serta sifat-sifat limit fungsi Dengan menggunakan sifat-sifat limit fungsi danrumus selisih cosinus tersebut, persoalan limit di atas dapat diselesaikan dengan cara sebagai berikut: Maka, . Oleh karena itu, jawaban yang benar adalah B.

Ingatlah rumus selisih cosinus

cos space straight A minus cos space straight B equals negative 2 space sin space 1 half open parentheses straight A plus straight B close parentheses space sin space 1 half open parentheses straight A minus straight B close parentheses

serta sifat-sifat limit fungsi

$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals a over b$
$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b$

Dengan menggunakan sifat-sifat limit fungsi dan rumus selisih cosinus tersebut, persoalan limit di atas dapat diselesaikan dengan cara sebagai berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses cos space 2 x minus cos space x close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses negative 2 space sin 1 half open parentheses 2 x plus x close parentheses space sin 1 half open parentheses 2 x minus x close parentheses close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator 2 space sin space begin display style 3 over 2 end style x space sin space 1 half x over denominator x space tan space x end fraction end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction close parentheses end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction end cell row blank equals cell 2 times open parentheses fraction numerator begin display style 3 over 2 end style over denominator 1 end fraction close parentheses times open parentheses fraction numerator begin display style 1 half end style over denominator 1 end fraction close parentheses end cell row blank equals cell 3 over 2 end cell row blank equals cell 1 comma 5 end cell end table$

Maka, $limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction equals 1 comma 5$ .

Oleh karena itu, jawaban yang benar adalah B.